Help me with my Calculus 1 problem. Here is a photo

Kent Beduya

Answered question

2022-09-01

nick1337

Expert2023-05-29Added 777 answers

To solve the given limit, we can directly substitute

x = - 1

into the expression

\frac{x^{2} - 1}{x^{2} + 4 x + 3}

and evaluate the result.
Let's substitute

x = - 1

into the expression:

{lim}_{x \to - 1} \frac{x^{2} - 1}{x^{2} + 4 x + 3} = \frac{(- 1)^{2} - 1}{(- 1)^{2} + 4 (- 1) + 3}

Simplifying the numerator and denominator, we have:

{lim}_{x \to - 1} \frac{1 - 1}{1 - 4 + 3}

{lim}_{x \to - 1} \frac{0}{0}

Here, we have an indeterminate form of

\frac{0}{0}

.
To evaluate this limit further, we can factor the numerator and denominator:

{lim}_{x \to - 1} \frac{(x - 1) (x + 1)}{(x + 1) (x + 3)}

Now, we can cancel out the common factor

(x + 1)

{lim}_{x \to - 1} \frac{x - 1}{x + 3}

Substituting

x = - 1

into this simplified expression, we have:

{lim}_{x \to - 1} \frac{- 1 - 1}{- 1 + 3}

{lim}_{x \to - 1} \frac{- 2}{2}

{lim}_{x \to - 1} - 1

Therefore, the limit as

x

approaches

- 1

\frac{x^{2} - 1}{x^{2} + 4 x + 3}

is equal to

- 1

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